Explanation

(b)
Since, $x^{2} + 20 + \sqrt{x^{4} + 20} = 22 + 20$
Let $\sqrt{x^{4} + 20} = y$
$\therefore y^{2} + y - 42 = 0$
$\Rightarrow \ \ (y - 6)(y + 7) = 0\ \Rightarrow y = 6\ \ \ \ \ (\because y \neq - 7)$
$\Rightarrow \ \ \sqrt{x^{4} + 20} = 6\ \ \ \ \Rightarrow \ \ \ \ x^{4} + 20 = 36$
$\Rightarrow \ \ \ \ x^{4} = 16\ \ \ \ \ \ \Rightarrow \ \ \ \ \ x = \pm 2$
Hence, the number of real roots of the equation is 2